Question

A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.380 m/s^2. Suppose it moves with constant acceleration for 3.80 s.
Required:
a. Find the position of the particle after this time.
b. Find its velocity at the end of this time interval.

Answer 1

Answer:

a

  $x_2 = -2.3356$

b

 $v = -1.384 \ m/s$

Explanation:

From the question we are told that

  The initial position of the particle is  $x_1 = 0.180 \ m$

  The initial  velocity of the particle is  $u = 0.060 \ m/s$

  The acceleration is   $a = -0.380 \ m/s^2$

   The time duration is  $t = 3.80 \ s$

Generally from kinematic equation

    $v = u + at$

=>  $v = 0.060 + (-0.380 * 3.80)$

=>  $v = -1.384 \ m/s$

Generally from kinematic equation

   $v^2 = u^2 + 2as$

Here s is the distance covered by the particle, so

   $(-1.384)^2 = (0.060)^2 + 2(-0.380)* s$

=>  $s = -2.5156 \ m$

Generally the final position of the particle is  

    $x_2 = x_1 + s$

=>   $x_2 = 0.180 + (-2.5156)$

=>   $x_2 = -2.3356$